Internal problem ID [6450]
Internal file name [OUTPUT/5698_Sunday_June_05_2022_03_47_35_PM_13412227/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 3(a).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{3} y^{\prime \prime }+\left (-1+\cos \left (2 x \right )\right ) y^{\prime }+2 y x=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{3} y^{\prime \prime }+\left (-1+\cos \left (2 x \right )\right ) y^{\prime }+2 y x = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {-1+\cos \left (2 x \right )}{x^{3}}\\ q(x) &= \frac {2}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{3} y^{\prime \prime }+\left (-1+\cos \left (2 x \right )\right ) y^{\prime }+2 y x = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-1+\cos \left (2 x \right )\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x = 0 \end{equation} Expanding \(-1+\cos \left (2 x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} -1+\cos \left (2 x \right ) &= -2 x^{2}+\frac {2}{3} x^{4}-\frac {4}{45} x^{6}+\frac {2}{315} x^{8} + \dots \\ &= -2 x^{2}+\frac {2}{3} x^{4}-\frac {4}{45} x^{6}+\frac {2}{315} x^{8} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {2 x^{n +r +7} a_{n} \left (n +r \right )}{315}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {4 x^{n +r +5} a_{n} \left (n +r \right )}{45}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {2 x^{n +r +3} a_{n} \left (n +r \right )}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(1+n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{1+n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\frac {2 x^{n +r +7} a_{n} \left (n +r \right )}{315} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {2 a_{n -6} \left (-6+n +r \right ) x^{1+n +r}}{315} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {4 x^{n +r +5} a_{n} \left (n +r \right )}{45}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {4 a_{n -4} \left (-4+n +r \right ) x^{1+n +r}}{45}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {2 x^{n +r +3} a_{n} \left (n +r \right )}{3} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {2 a_{n -2} \left (n +r -2\right ) x^{1+n +r}}{3} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(1+n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {2 a_{n -6} \left (-6+n +r \right ) x^{1+n +r}}{315}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {4 a_{n -4} \left (-4+n +r \right ) x^{1+n +r}}{45}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {2 a_{n -2} \left (n +r -2\right ) x^{1+n +r}}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 x^{1+n +r} a_{n} \left (n +r \right )+2 x^{1+n +r} a_{n} = 0 \] When \(n = 0\) the above becomes \[ x^{1+r} a_{0} r \left (-1+r \right )-2 x^{1+r} a_{0} r +2 x^{1+r} a_{0} = 0 \] Or \[ \left (x^{1+r} r \left (-1+r \right )-2 x^{1+r} r +2 x^{1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-3 r +2\right ) x^{1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-3 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-3 r +2\right ) x^{1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, 1]\).
Since \(r_1 - r_2 = 1\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{1+n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = -\frac {2}{3 r +3} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = 0 \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {2 a_{n -6} \left (-6+n +r \right )}{315}-\frac {4 a_{n -4} \left (-4+n +r \right )}{45}+\frac {2 a_{n -2} \left (n +r -2\right )}{3}-2 a_{n} \left (n +r \right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 \left (n a_{n -6}-14 n a_{n -4}+105 n a_{n -2}+r a_{n -6}-14 r a_{n -4}+105 r a_{n -2}-6 a_{n -6}+56 a_{n -4}-210 a_{n -2}\right )}{315 \left (n^{2}+2 n r +r^{2}-3 n -3 r +2\right )}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {\left (-2 a_{n -6}+28 a_{n -4}-210 a_{n -2}\right ) n +8 a_{n -6}-56 a_{n -4}}{315 n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {2}{9}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {26}{675}\) |
| \(a_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-\frac {2}{315} r^{4}-\frac {148}{945} r^{3}-\frac {1018}{945} r^{2}-\frac {2836}{945} r -\frac {416}{135}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right ) \left (5+r \right ) \left (4+r \right )} \] Which for the root \(r = 2\) becomes \[ a_{6}=-{\frac {1742}{297675}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {2}{9}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {26}{675}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(\frac {-\frac {2}{315} r^{4}-\frac {148}{945} r^{3}-\frac {1018}{945} r^{2}-\frac {2836}{945} r -\frac {416}{135}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right ) \left (5+r \right ) \left (4+r \right )}\) | \(-{\frac {1742}{297675}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {2}{9}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {26}{675}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(\frac {-\frac {2}{315} r^{4}-\frac {148}{945} r^{3}-\frac {1018}{945} r^{2}-\frac {2836}{945} r -\frac {416}{135}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right ) \left (5+r \right ) \left (4+r \right )}\) | \(-{\frac {1742}{297675}}\) |
| \(a_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{2} \left (1-\frac {2 x^{2}}{9}+\frac {26 x^{4}}{675}-\frac {1742 x^{6}}{297675}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= 0 \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow 1}0\\ &= 0 \end {align*}
The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{1+n} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = 0 \] Substituting \(n = 2\) in Eq(3) gives \[ b_{2} = -\frac {2}{3 \left (1+r \right )} \] Substituting \(n = 3\) in Eq(3) gives \[ b_{3} = 0 \] Substituting \(n = 4\) in Eq(3) gives \[ b_{4} = \frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (r^{2}+5 r +6\right )} \] Substituting \(n = 5\) in Eq(3) gives \[ b_{5} = 0 \] For \(6\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {2 b_{n -6} \left (-6+n +r \right )}{315}-\frac {4 b_{n -4} \left (-4+n +r \right )}{45}+\frac {2 b_{n -2} \left (n +r -2\right )}{3}-2 b_{n} \left (n +r \right )+2 b_{n} = 0 \end{equation} Which for for the root \(r = 1\) becomes \begin{equation} \tag{4A} b_{n} \left (1+n \right ) n +\frac {2 b_{n -6} \left (-5+n \right )}{315}-\frac {4 b_{n -4} \left (-3+n \right )}{45}+\frac {2 b_{n -2} \left (n -1\right )}{3}-2 b_{n} \left (1+n \right )+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {2 \left (n b_{n -6}-14 n b_{n -4}+105 n b_{n -2}+r b_{n -6}-14 r b_{n -4}+105 r b_{n -2}-6 b_{n -6}+56 b_{n -4}-210 b_{n -2}\right )}{315 \left (n^{2}+2 n r +r^{2}-3 n -3 r +2\right )}\tag {5} \] Which for the root \(r = 1\) becomes \[ b_{n} = -\frac {2 \left (n b_{n -6}-14 n b_{n -4}+105 n b_{n -2}-5 b_{n -6}+42 b_{n -4}-105 b_{n -2}\right )}{315 \left (n^{2}-n \right )}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {17}{270}\) |
| \(b_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=-\frac {2 \left (3 r^{4}+74 r^{3}+509 r^{2}+1418 r +1456\right )}{945 \left (1+r \right ) \left (r^{2}+5 r +6\right ) \left (r^{2}+9 r +20\right )} \] Which for the root \(r = 1\) becomes \[ b_{6}=-{\frac {173}{17010}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {17}{270}\) |
| \(b_{5}\) | \(0\) | \(0\) |
| \(b_{6}\) | \(\frac {-\frac {2}{315} r^{4}-\frac {148}{945} r^{3}-\frac {1018}{945} r^{2}-\frac {2836}{945} r -\frac {416}{135}}{\left (1+r \right ) \left (r^{2}+5 r +6\right ) \left (r^{2}+9 r +20\right )}\) | \(-{\frac {173}{17010}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {2}{3 r +3}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\frac {4}{45} r^{2}+\frac {8}{15} r +\frac {8}{9}}{\left (1+r \right ) \left (3+r \right ) \left (2+r \right )}\) | \(\frac {17}{270}\) |
| \(b_{5}\) | \(0\) | \(0\) |
| \(b_{6}\) | \(\frac {-\frac {2}{315} r^{4}-\frac {148}{945} r^{3}-\frac {1018}{945} r^{2}-\frac {2836}{945} r -\frac {416}{135}}{\left (1+r \right ) \left (r^{2}+5 r +6\right ) \left (r^{2}+9 r +20\right )}\) | \(-{\frac {173}{17010}}\) |
| \(b_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{2} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= x \left (1-\frac {x^{2}}{3}+\frac {17 x^{4}}{270}-\frac {173 x^{6}}{17010}+O\left (x^{8}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{2} \left (1-\frac {2 x^{2}}{9}+\frac {26 x^{4}}{675}-\frac {1742 x^{6}}{297675}+O\left (x^{8}\right )\right ) + c_{2} x \left (1-\frac {x^{2}}{3}+\frac {17 x^{4}}{270}-\frac {173 x^{6}}{17010}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1-\frac {2 x^{2}}{9}+\frac {26 x^{4}}{675}-\frac {1742 x^{6}}{297675}+O\left (x^{8}\right )\right )+c_{2} x \left (1-\frac {x^{2}}{3}+\frac {17 x^{4}}{270}-\frac {173 x^{6}}{17010}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1-\frac {2 x^{2}}{9}+\frac {26 x^{4}}{675}-\frac {1742 x^{6}}{297675}+O\left (x^{8}\right )\right )+c_{2} x \left (1-\frac {x^{2}}{3}+\frac {17 x^{4}}{270}-\frac {173 x^{6}}{17010}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1-\frac {2 x^{2}}{9}+\frac {26 x^{4}}{675}-\frac {1742 x^{6}}{297675}+O\left (x^{8}\right )\right )+c_{2} x \left (1-\frac {x^{2}}{3}+\frac {17 x^{4}}{270}-\frac {173 x^{6}}{17010}+O\left (x^{8}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.235 (sec). Leaf size: 37
Order:=8; dsolve(x^3*diff(y(x),x$2)+(cos(2*x)-1)*diff(y(x),x)+2*x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1-\frac {2}{9} x^{2}+\frac {26}{675} x^{4}-\frac {1742}{297675} x^{6}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x \left (1-\frac {1}{3} x^{2}+\frac {17}{270} x^{4}-\frac {173}{17010} x^{6}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 74
AsymptoticDSolveValue[x^3*y''[x]+(Cos[2*x]-1)*y'[x]+2*x*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {32351 x^8}{40186125}-\frac {1742 x^6}{297675}+\frac {26 x^4}{675}-\frac {2 x^2}{9}+1\right ) x^2+c_1 \left (\frac {10471 x^8}{7144200}-\frac {173 x^6}{17010}+\frac {17 x^4}{270}-\frac {x^2}{3}+1\right ) x \]